题目内容
计算:
(1)
÷(1-
)
(2)
÷(a+2)×
(3)(
-
)÷
(4)
-
×
.
(1)
| x-2 |
| x2 |
| 2 |
| x |
(2)
| 3a+6 |
| a2-6a+9 |
| 9-a2 |
| (a+2)(a+3) |
(3)(
| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| x-4 |
| x |
(4)
| 1 |
| x-1 |
| x-3 |
| x2-1 |
| x2+2x+1 |
| x2-6x+9 |
考点:分式的混合运算
专题:
分析:(1)先算括号里面的,再将除法转化为乘法计算;
(2)将分子分母因式分解,再将除法转化为乘法解答;
(3)将括号内的部分通分后相加,再将除法转化为乘法解答;
(4)将分子分母因式分解,约分后相减即可.
(2)将分子分母因式分解,再将除法转化为乘法解答;
(3)将括号内的部分通分后相加,再将除法转化为乘法解答;
(4)将分子分母因式分解,约分后相减即可.
解答:解:(1)原式=
÷
=
•
=
;
(2)原式=
•
•
=-
;
(3)(
-
)÷
=[
-
]•
=[
-
]•
=
•
=
•
=
;
(4)
-
×
=
-
×
=
-
=
-
=0.
| x-2 |
| x2 |
| x-2 |
| x |
=
| x-2 |
| x2 |
| x |
| x-2 |
=
| 1 |
| x |
(2)原式=
| 3(a+2) |
| (a-3)2 |
| 1 |
| a+2 |
| (3-a)(3+a) |
| (a+2)(a+3) |
=-
| 3 |
| a2-a-6 |
(3)(
| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| x-4 |
| x |
=[
| x+2 |
| x(x-2) |
| x-1 |
| (x-2)2 |
| x |
| x-4 |
=[
| (x+2)(x-2) |
| x(x-2)2 |
| x2-x |
| x(x-2)2 |
| x |
| x-4 |
=
| x2-4-x2+x |
| x(x-2)2 |
| x |
| x-4 |
=
| x-4 |
| x(x-2)2 |
| x |
| x-4 |
=
| x |
| x2-4x+4 |
(4)
| 1 |
| x-1 |
| x-3 |
| x2-1 |
| x2+2x+1 |
| x2-6x+9 |
=
| 1 |
| x-1 |
| x-3 |
| (x-1)(x+1) |
| (x+1)2 |
| (x-3)2 |
=
| 1 |
| x-1 |
| x+1 |
| (x-1)(x-3) |
=
| x+1 |
| (x-1)(x-3) |
| x+1 |
| (x-1)(x-3) |
=0.
点评:本题考查了分式的混合运算,熟悉因式分解、熟悉分式的乘除法是解题的关键.
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